cookie monster meets the fibonacci numbers. mmmmmm theorems! · intro generalizations gaps more...

Intro Generalizations Gaps More Gaps

Cookie Monster Meets the FibonacciNumbers. Mmmmmm – Theorems!

Louis Gaudet (advisor Steven J Miller)http://www.williams.edu/Mathematics/sjmiller/public html

SUMS, Providence, RI, March 10, 2012

1


Introduction

2


Goals of the Talk

Some linear recursions and decompositions.

Uncover some of the secrets of gaps.

Methods: Combinatorial vantage, Binet-like formulas.

Specific open problems.

Thanks to my advisor and his colleagues from the WilliamsCollege 2010 and 2011 SMALL REU programs.

3


Previous Results

Fibonacci Numbers: Fn+1 = Fn + Fn−1;F1 = 1, F2 = 2, F3 = 3, F4 = 5, . . . .

4


Previous Results


Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum ofnon-consecutive Fibonacci numbers.

5


Previous Results



Example:2012 = 1597 + 377 + 34 + 3 + 1 = F16 + F13 + F8 + F3 + F1.

6


Previous Results



Example:2012 = 1597 + 377 + 34 + 3 + 1 = F16 + F13 + F8 + F3 + F1.

Lekkerkerker’s Theorem (1952)

The average number of summands in the Zeckendorfdecomposition for integers in [Fn,Fn+1) tends to n

ϕ2+1 ≈ .276n,

where ϕ = 1+√

52 is the golden mean.

7


Previous Results

Central Limit Type Theorem

As n → ∞, the distribution of the number of summands in theZeckendorf decomposition for integers in [Fn,Fn+1) is Gaussian(normal).

500 520 540 560 580 600

0.005

0.010

0.015

0.020

0.025

0.030

Figure: Number of summands in [F2010,F2011); F2010 ≈ 10420.8


New Results

Theorem (Zeckendorf Gap Distribution (BM))

For Zeckendorf decompositions, P(k) = ϕ(ϕ−1)ϕk for k ≥ 2, with

ϕ = 1+√

52 the golden mean.

5 10 15 20 25 30

0.1

0.2

0.3

0.4

5 10 15 20 25

0.5

1.0

1.5

2.0

Figure: Distribution of gaps in [F1000,F1001); F2010 ≈ 10208.

9


Preliminaries: The Cookie Problem

The Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is

(C+P−1P−1

)

.

10




(C+P−1P−1

)

.

Proof : Consider C + P − 1 cookies in a line.Cookie Monster eats P − 1 cookies:

(C+P−1P−1

)

ways to do.Divides the cookies into P sets.

11




(C+P−1P−1

)

.


(C+P−1P−1

)

ways to do.Divides the cookies into P sets.Example: 8 cookies and 5 people (C = 8, P = 5):

12




(C+P−1P−1

)

.


(C+P−1P−1

)


13




(C+P−1P−1

)

.


(C+P−1P−1

)


14


Preliminaries: The Cookie Problem: Reinterpretation

Reinterpreting the Cookie Problem

The number of solutions to x1 + · · · + xP = C with xi ≥ 0 is(C+P−1

P−1

)

.

15





P−1

)

.

Let pn,k = # {N ∈ [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.

16





P−1

)

.

Let pn,k = # {N ∈ [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.

For N ∈ [Fn,Fn+1), the largest summand is Fn.

N = Fi1 + Fi2 + · · · + Fik−1+ Fn,

1 ≤ i1 < i2 < · · · < ik−1 < ik = n, ij − ij−1 ≥ 2.

d1 := i1 − 1, dj := ij − ij−1 − 2 (j > 1).

d1 + d2 + · · ·+ dk = n − 2k + 1, dj ≥ 0.

Cookie counting ⇒ pn,k =(n−2k+1−k−1

k−1

)

=(n−k

k−1

)

.

17


Generalizations

18


Generalizations

Generalizing from Fibonacci numbers to linearly recursivesequences with arbitrary nonnegative coefficients.

Hn+1 = c1Hn + c2Hn−1 + · · · + cLHn−L+1, n ≥ L

with H1 = 1, Hn+1 = c1Hn + c2Hn−1 + · · ·+ cnH1 + 1, n < L,coefficients ci ≥ 0; c1, cL > 0 if L ≥ 2; c1 > 1 if L = 1.

Zeckendorf: Every positive integer can be written uniquelyas∑

aiHi with natural constraints on the ai ’s(e.g. cannot use the recurrence relation to remove anysummand).

Lekkerkerker


19


Generalizing Lekkerkerker

Generalized Lekkerkerker’s TheoremThe average number of summands in the generalizedZeckendorf decomposition for integers in [Hn,Hn+1) tends toCn + d as n → ∞, where C > 0 and d are computableconstants determined by the ci ’s.

C = −y ′(1)y(1)

=

∑L−1m=0(sm + sm+1 − 1)(sm+1 − sm)ym(1)

2∑L−1

m=0(m + 1)(sm+1 − sm)ym(1).

s0 = 0, sm = c1 + c2 + · · ·+ cm.

y(x) is the root of 1 −∑L−1

m=0

∑sm+1−1j=sm

x jym+1.

y(1) is the root of 1 − c1y − c2y2 − · · · − cLyL.

20




As n → ∞, the distribution of the number of summands, i.e.,a1 + a2 + · · ·+ am in the generalized Zeckendorf decomposition∑m

i=1 aiHi for integers in [Hn,Hn+1) is Gaussian.

1000 1050 1100 1150 1200

0.005

0.010

0.015

0.020

21


Gaps Between Summands

22


Distribution of Gaps

For Fi1 + Fi2 + · · ·+ Fin , the gaps are the differencesin − in−1, in−1 − in−2, . . . , i2 − i1.

23




Example: For F1 + F8 + F18, the gaps are 7 and 10.

24





Let Pn(k) be the probability that a gap for a decomposition in[Fn,Fn+1) is of length k .

25






What is P(k) = limn→∞ Pn(k)?

26






What is P(k) = limn→∞ Pn(k)?

Can ask similar questions about binary or other expansions:2011 = 210 + 29 + 28 + 27 + 26 + 24 + 23 + 21 + 20.

27


Main Results (Beckwith-Miller 2011)

Theorem (Base B Gap Distribution)

For base B decompositions, P(0) = (B−1)(B−2)B2 , and for k ≥ 1,

P(k) = cBB−k , with cB = (B−1)(3B−2)B2 .

Theorem (Zeckendorf Gap Distribution)

For Zeckendorf decompositions, P(k) = ϕ(ϕ−1)ϕk for k ≥ 2, with

ϕ = 1+√

52 the golden mean.

28


Proof of Fibonacci Result

29



Let Xi ,j(n) = #{m ∈ [Fn,Fn+1): decomposition of m includes Fi ,Fj , but not Fq for i < q < j}.

30




Let Y (n) = total number of gaps in decompositions for integersin [Fn,Fn+1)

31





P(k) = limn→∞

1Y (n)

n−k∑

i=1

Xi ,i+k(n).

32





P(k) = limn→∞

1Y (n)

n−k∑

i=1

Xi ,i+k(n).

Lekkerkerker ⇒ Y (n) ∼ Fn−1n

ϕ2+1 .

33


Calculating Xi ,i+k

34


Calculating Xi ,i+k

In the interval [Fn,Fn+1):How many decompositions contain a gap from Fi to Fi+k?

35


Calculating Xi ,i+k


For the indices less than i : Fi−1 choices.

36


Calculating Xi ,i+k



For the indices greater than i + k : Fn−k−i−2 choices.

37


Calculating Xi ,i+k




So total choices number of choices is Fi−1Fn−k−i−2.

38


Calculating Xi ,i+k




So total choices number of choices is Fi−1Fn−k−i−2.

Xi ,i+k(n) = Fi−1Fn−k−i−2

39


Finding P(k)

40


Finding P(k)

Binet-like Formula : Fn = c1rn1 + c2rn

2

41


Finding P(k)


2

c1 =5 +

√5

10, c2 = c̄1 and r1 = ϕ =

1 +√

52

, r2 = r̄1

42


Finding P(k)


2

c1 =5 +

√5

10, c2 = c̄1 and r1 = ϕ =

1 +√

52

, r2 = r̄1

Xi ,i+k(n) = Fi−1Fn−k−i−2 = c21rn−k−3

1 + smaller

43


Finding P(k)


2

c1 =5 +

√5

10, c2 = c̄1 and r1 = ϕ =

1 +√

52

, r2 = r̄1

Xi ,i+k(n) = Fi−1Fn−k−i−2 = c21rn−k−3

1 + smaller

Y (n) = Fn−1n

ϕ2 + 1+ smaller =

nϕ2 + 1

c1rn−11 + smaller

44


Finding P(k)

P(k) = limn→∞

1Y (n)

n−k∑

i=1

Xi ,i+k(n)

= limn→∞

∑n−ki=1 c2

1rn−k−31

nϕ2+1c1rn−1

1

=ϕ(ϕ − 1)

ϕk for k ≥ 2

45


Gaps for other Linear Recurrences

46


Tribonacci Gaps

Tribonacci Numbers: Tn+1 = Tn + Tn−1 + Tn−2;F1 = 1, F2 = 2, F3 = 4, F4 = 7, . . . .

47


Tribonacci Gaps


Consider the interval [Tn,Tn+1):

48


Tribonacci Gaps



Generalized Lekkerkerker:

Y (n) = Cn(Tn−1 + Tn−2) + smaller

49


Tribonacci Gaps



Generalized Lekkerkerker:

Y (n) = Cn(Tn−1 + Tn−2) + smaller

Counting:

Xi ,i+k(n) ={

Ti−1(Tn−i−3 + Tn−i−4) if k = 1(Ti−1 + Ti−2)(Tn−k−i−1 + Tn−k−i−3) if k ≥ 2

50


Tribonacci Gaps

P(k) = limn→∞1

Y (n)

∑n−ki=1 Xi ,i+k(n)

51


Tribonacci Gaps

P(k) = limn→∞1

Y (n)


Closed form: Tn = c1λn1 + c2λ

n2 + c3λ

n3, |λ1| > |λ2| = |λ3|

52


Tribonacci Gaps

P(k) = limn→∞1

Y (n)



n2 + c3λ

n3, |λ1| > |λ2| = |λ3|

P(1) =c1

Cλ31

53


Tribonacci Gaps

P(k) = limn→∞1

Y (n)



n2 + c3λ

n3, |λ1| > |λ2| = |λ3|

P(1) =c1

Cλ31

P(k) =2c1

C(1 + λ1)λ−k

1 (for k ≥ 2)

54


Tribonacci Gaps

P(k) = limn→∞1

Y (n)



n2 + c3λ

n3, |λ1| > |λ2| = |λ3|

P(1) =c1

Cλ31

P(k) =2c1

C(1 + λ1)λ−k

1 (for k ≥ 2)

∑∞k=1 P(k) = 1 ⇒ C = c1

(

3λ21 − 1

(λ21 − 1)λ3

1

)

55


Other gaps?

Gaps longer than recurrence – should be geometric decay

56


Other gaps?


Interesting behavior with “short” gaps

57


Other gaps?



“Skiponaccis”: Sn+1 = Sn + Sn−2

58


Other gaps?




“Doublanaccis”: Hn+1 = 2Hn + Hn−1

59


Other gaps?




“Doublanaccis”: Hn+1 = 2Hn + Hn−1

Generalize to all positive linear recurrences?

60

cookie monster meets the fibonacci numbers. mmmmmm theorems! · intro generalizations gaps more...

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